JPA LazyInitialisationException - 无法初始化代理

问题描述 投票:0回答:1

是的,我已经看到了类似的问题,但没有答案竟使我的解决方案。我不使用线程的任何地方,而这种相同的代码工作在另一个jhipster的应用程序,所以我难倒,为什么这个电梯+转变使得Hibernate的问题。

public UserDetailsContextMapper userDetailsContextMapper() {
    return new UserDetailsContextMapper() {
        @Override
        public UserDetails mapUserFromContext(
            DirContextOperations ctx, String username,
            Collection<? extends GrantedAuthority> authorities) {
            log.error("2 " + username + " -> " + ctx.toString());
            String lowercaseLogonName = username.toLowerCase();
            Optional<User> userFromDatabase = userRepository.findOneByLogin(lowercaseLogonName);
            if (!userFromDatabase.isPresent()) {
                User ldapUser = new User();
                ldapUser.setLogin(lowercaseLogonName);
                ldapUser.setPassword(RandomStringUtils.random(60)); // We use LDAP password, but the password need to be set
                ldapUser.setActivated(true);
                try {
                    Attribute attribute = ctx.getAttributes().get("mail");
                    ldapUser.setEmail( attribute != null && attribute.size() > 0 ? (String) attribute.get(0) : "");
                    attribute = ctx.getAttributes().get("givenname");
                    ldapUser.setFirstName(attribute != null && attribute.size() > 0 ? (String) attribute.get(0) : "");
                    attribute = ctx.getAttributes().get("sn");
                    ldapUser.setLastName(attribute != null && attribute.size() > 0 ? (String) attribute.get(0) : "");
                } catch (NamingException e) {
                    log.warn("Couldn't get LDAP details for user: " + lowercaseLogonName);
                }

                Set<Authority> newUserAuthorities = new HashSet<>();
                Authority authority = authorityRepository.getOne(AuthoritiesConstants.USER);

                newUserAuthorities.add(authority); <-- This line in the Exception

                ldapUser.setAuthorities(newUserAuthorities);
                ldapUser.setLangKey("en");
                userRepository.saveAndFlush(ldapUser);

            }

例外:

org.hibernate.LazyInitializationException: could not initialize proxy [com.app.my.let.domain.Authority#ROLE_USER] - no Session
        at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:155)
        at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:268)
        at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:73)
        at com.app.my.let.domain.Authority_$$_jvst758_1e.hashCode(Authority_$$_jvst758_1e.java)
        at java.util.HashMap.hash(HashMap.java:339)
        at java.util.HashMap.put(HashMap.java:612)
        at java.util.HashSet.add(HashSet.java:220)
        at com.app.my.let.config.SecurityConfiguration$1.mapUserFromContext(SecurityConfiguration.java:249) <-- code line above

JPA属性:

jpa:
        database-platform: org.hibernate.dialect.SQLServer2012Dialect
        database: SQL_SERVER
        show_sql: true
        format_sql: true
        properties:
            hibernate.id.new_generator_mappings: true
            hibernate.cache.use_second_level_cache: false
            hibernate.cache.use_query_cache: false
            hibernate.generate_statistics: true

编辑 - 这解决了我的问题:

Optional<Authority> optAuthority = authorityRepository.findById(AuthoritiesConstants.USER);

Authority authority = optAuthority.get();
java spring hibernate jpa jhipster
1个回答
2
投票

这条线:

Authority authority = authorityRepository.getOne(AuthoritiesConstants.USER);

给你刚刚提到的实体。引擎盖下它调用EntityManager#getReference。作为赛斯,这不是实体,只是对象关系映射目的的参考。该异常被抛出,因为你试图投它事务环境之外。

有几个解决方案。要获得实体,你将不得不使用EntityManager#find或做事务中(使用@Transactional在法)。在这里,你还可以调用Hibernate.initialize(authority)从数据库中读取实体。

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